limn→∞∑k=1n(2+3kn−−−−−−√4⋅3n)=
∫2 to 5 4√xⅆx
To evaluate the limit, we need to find the behavior of the given expression as n approaches infinity. We can start by factoring out the constant term 2 from the square root to get:
lim n→∞ ∑k=1n 2(1 + 3k/n)^1/2 / 2ᵏ³
Next, we notice that the expression inside the summation looks like a Riemann sum with Δx = 1/n and xᵢ = 3k/n for i = 1, 2,…, n. Therefore, we can replace the summation with an integral from 0 to 3:
lim n→∞ ∫0³ 2(1 + 3x)^1/2 / 2³x³ dx
Simplifying the expression above, we get:
lim n→∞ 1/16 ∫0³ (1 + 3x)^1/2 / x³ dx
We can evaluate the integral by using substitution u = 1 + 3x and du = 3dx. This gives:
lim n→∞ 1/16 ∫1⁹ u^1/2 / (u – 1)³ du
We can now use partial fractions to expand the integrand as follows:
u^1/2 / (u – 1)³ = A/(u-1) + B/(u-1)^2 + C/(u-1)^3
Solving for the constants A, B, and C, we get:
A = 1/2
B = 3/4
C = -1/4
Substituting these values back, we get:
lim n→∞ 1/16 [(1/2)ln|u-1| – (3/4)/(u-1) – (1/4)/(u-1)^2]₁ ⁹
Evaluating the limits at both ends, we get:
lim n→∞ 1/16 [(1/2)ln(8) – (3/4)/(8) – (1/4)/(8)^2 – (1/2)ln(2) + 3/4 – 1/32]
Simplifying this expression, we get:
lim n→∞ [ln(2) + 9/16 – 1/32]
Therefore, the limit evaluates to:
ln(2) + 17/32
Conclusion:
limn→∞∑k=1n(2+3kn−−−−−−√4⋅3n) = ln(2) + 17/32.
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